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30 Jan 2017, 16:40
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
42% (01:33) correct
58%
(01:13)
wrong
based on 233
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How many different solutions the equation (x^4 - 1)/(x^2 - 1) = 2 have?
a) 0
b) 1
c) 2
d) 3
e) 4
oa:
I keep getting
a) 0
b) 1
c) 2
d) 3
e) 4
oa:
I keep getting
30 Jan 2017, 19:34
Kudos
Bookmarks
robypeter
How many different solutions the equation {x^4-1/}{x^2-1}= 2 have?
a) 0 b) 1 c) 2 d) 3 e) 4
oa:
I keep getting
a) 0 b) 1 c) 2 d) 3 e) 4
oa:
I keep getting
Hi,
If you substitute the values you will find the solutions are not valid
\(\frac{x^4-1}{x^2-1}\)=2....
\(\frac{(x^2-1)(x^2+1)}{x^2-1}\)=2..
So what you are doing now is, cut x^2-1 from both and x^2+1=2 or x=1 or x=-1..
But if you substitute this in denominator x^2-1, it becomes 0 and thus the equation becomes invalid..
So ZERO solutions.
A
30 Jan 2017, 22:38
Kudos
Bookmarks
robypeter
How many different solutions the equation {x^4-1/}{x^2-1}= 2 have?
a) 0 b) 1 c) 2 d) 3 e) 4
oa:
I keep getting
a) 0 b) 1 c) 2 d) 3 e) 4
oa:
I keep getting
Since x^2 - 1 is in the denominator, it cannot be 0.
So x^2 is not 1 i.e. x is not 1 or -1.
Now cross multiply to get
\((x^4 - 1) = 2*(x^2 - 1)\)
\(x^4 - 2x^2 + 1 = 0\)
\((x^2 - 1)^2 = 0\)
x^2 - 1 = 0
x is 1 or -1.
But x cannot take both those values. Hence answer is (A).
Always remember, when you have a denominator, step 1 should be that it cannot be 0. Often, it will be given in the question itself. If it is not given, write it down.
